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A box contains 20 electric bulbs, out of which 4 are defective. Two

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A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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New post 22 Nov 2011, 22:30
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A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are drawn at random . The probability that atleast one of these is defective

1) 4/19
2) 7/19
3) 12/19
4) 10/19


for atleast 1 , i solved it like this

4c1+4c2 /20c2 and i get 1/19 as the answer but the answer is 7/19.
Pls lemme know where i am going wrong.
[Reveal] Spoiler: OA

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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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New post 22 Nov 2011, 23:31
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vaibhav123 wrote:
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are drawn at random . The probability that atleast one of these is defective

1)4/19
2)7/19
3)12/19
4)10/19


for atleast 1 , i solved it like this

4c1+4c2 /20c2 and i get 1/19 as the answer but the answer is 7/19.
Pls lemme know where i am going wrong.


Remember one thing -for all the question having "atleast" option, try to find the probablity in reverse way.
That means -
P(probability that atleast one of the bulb is defective) = 1 - P(probability that none of the bulb is defective)
= 1 - (16C2/20C2) = 7/19.
Hope it helps. Cheers!

P.S. - 4c1+4c2 - Can't be used because the question doesn't say that bulbs picked one by one.
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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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New post 23 Nov 2011, 03:14
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vaibhav123 wrote:
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are drawn at random . The probability that atleast one of these is defective

1)4/19
2)7/19
3)12/19
4)10/19


for atleast 1 , i solved it like this

4c1+4c2 /20c2 and i get 1/19 as the answer but the answer is 7/19.
Pls lemme know where i am going wrong.


Your answer is almost right. When you pick one defective bulb and one non-defective bulb, you need to count the number of ways to pick the non-defective bulb as well. So you can pick 1 bulb of each type in (4C1)(16C1) ways, and the answer is

[(4C1)(16C1) + 4C2 ] / 20C2

which is equal to 7/19.


Capricorn369 wrote:

Can't be used because the question doesn't say that bulbs picked one by one.


This actually makes no difference at all. After all, if you put your two hands in the box and take hold of two lightbulbs, it won't be any more or less likely you get two defective bulbs if you take your hands out of the box one at a time than if you take them both out at once.

I find it much easier in these types of problems to imagine making the selections one at a time. You can then avoid using expressions like '20C2' altogether. If I pick one bulb at a time, the probability the first is defective is 16/20, and the probability the second is then also defective is 15/19, so the probability both are defective is

(16/20)(15/19) = 12/19

and the probability at least one is defective is 1 - 12/19 = 7/19.
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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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New post 23 Nov 2011, 06:29
Your answer is almost right. When you pick one defective bulb and one non-defective bulb, you need to count the number of ways to pick the non-defective bulb as well. So you can pick 1 bulb of each type in (4C1)(16C1) ways, and the answer is

[(4C1)(16C1) + 4C2 ] / 20C2

which is equal to 7/19.



Good explanation ..thnks...:)

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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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New post 13 Dec 2011, 04:06
I prefer \(Capricorn369\)'s approach. But for ones who want another approach:

P1=probability that the first is defective and second is not= \(\frac{4}{20}*\frac{16}{19}\)

P2=probability that the first is intact and second is defective= \(\frac{16}{20}*\frac{4}{19}\)

P3=probability that both are defective= \(\frac{4}{20}*\frac{3}{19}\)

\(P=P1+P2+P3=\frac{7}{19}\)

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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two [#permalink]

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Re: A box contains 20 electric bulbs, out of which 4 are defective. Two   [#permalink] 28 Aug 2017, 23:01
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